- Set the two equations equal to each other: \(x^2 - 2x - 15 = 8x - 40\).
- Simplify and rewrite the equation in standard form: \(x^2 - 10x + 25 = 0\).
- Factor the quadratic as \((x - 5)^2 = 0\) to find the solution \(x = 5\).
- Substitute \(x = 5\) back into the linear equation to find \(y = 0\), resulting in the final solution \(\boxed{(5, 0)}\).
Explanation
- Problem Analysis and Strategy
We are given a system of two equations: \(y = x^2 - 2x - 15\) and \(y = 8x - 40\). Our goal is to find the point or points \((x, y)\) that satisfy both equations simultaneously. Since both expressions are set equal to \(y\), we can equate the two expressions for \(x\) to solve for its value.
- Solving for x
Setting the two expressions for \(y\) equal to each other, we have:
\[x^2 - 2x - 15 = 8x - 40\]
To solve for \(x\), we rearrange the equation to one side to form a standard quadratic equation of the form \(ax^2 + bx + c = 0\):
\[x^2 - 2x - 8x - 15 + 40 = 0\]
\[x^2 - 10x + 25 = 0\]
We can now factor the quadratic expression. Recognizing that \(x^2 - 10x + 25\) is a perfect square trinomial, we can factor it as \((x - 5)^2 = 0\).
Thus, the only solution for \(x\) is \(x = 5\).
- Solving for y and Final Conclusion
Now that we have the value of \(x\), we substitute it into the linear equation \(y = 8x - 40\) to find the corresponding \(y\)-value:
\[y = 8(5) - 40\]
\[y = 40 - 40\]
\[y = 0\]
Checking this result with the first equation:
\[y = 5^2 - 2(5) - 15 = 25 - 10 - 15 = 0\]
Since the value of \(y\) is consistent in both equations, the only solution to the system is the point \((5, 0)\). Comparing this to the given options, we select the correct one.
Examples
In engineering and economics, finding where two functions intersect—like supply and demand curves—is essential for determining equilibrium points. Just as we found where a parabola and a line meet at \((5, 0)\), professionals often identify these intersection points to optimize resource allocation, balance budgets, or predict physical outcomes in structural mechanics. This algebraic technique provides a robust way to solve for critical operational values.